What is the time complexity of Reward Top K Students?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Reward Top K Students?

This problem uses Array, Hash Table, String, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Reward Top K Students?

Insight 1: Representing positive and negative words efficiently using hash sets (or sets in Python) allows for quick lookup when calculating the score for each student. Insight 2: Storing student information (score and ID) as a pair (or a similar data structure) is crucial for sorting them based on score and ID.

Which companies ask LeetCode 2512. Reward Top K Students in interviews?

This problem is frequently asked at Booking.com.

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Reward Top K Students - LeetCode 2512 Solution

Reward Top K Students - Complete Solution Guide

Reward Top K Students is LeetCode problem 2512, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two string arrays positive_feedback and negative_feedback , containing the words denoting positive and negative feedback, respectively. Note that no word is both positive and negative. Initially every student has 0 points. Each positive word in a feedback report increases the points of a student by 3 , whereas each negative word decreases the points by 1 . You are given n feedback reports, represented by a 0-indexed string array report and a 0-indexed integer array student_id , whe

Detailed Explanation

The problem requires ranking students based on feedback they receive. Each student has a unique ID, and receives a feedback report consisting of words. A positive word increases their score by 3, while a negative word decreases it by 1. The goal is to return the IDs of the top `k` students, ranked in non-increasing order of their scores. If two students have the same score, the student with the lower ID ranks higher.

Solution Approach

The solution calculates each student's score by iterating through their feedback report, checking if each word is in the set of positive or negative words, and updating the score accordingly. It then stores the score and ID of each student in a list of pairs. This list is sorted based on the score (in descending order) and, if the scores are equal, based on the student ID (in ascending order). Finally, the first `k` student IDs from the sorted list are returned.

Step-by-Step Algorithm

Step 1: Create hash sets (or sets) for positive and negative feedback words for efficient lookup.
Step 2: Iterate through each student's report and calculate their score. For each word in the report, check if it's in the positive or negative feedback set and update the score accordingly.
Step 3: Store each student's score and ID as a pair in a list.
Step 4: Sort the list of student information based on the score in descending order (using a negative score for simpler sorting) and then by student ID in ascending order (for tie-breaking).
Step 5: Extract the top `k` student IDs from the sorted list and return them.

Key Insights

Insight 1: Representing positive and negative words efficiently using hash sets (or sets in Python) allows for quick lookup when calculating the score for each student.
Insight 2: Storing student information (score and ID) as a pair (or a similar data structure) is crucial for sorting them based on score and ID.
Insight 3: Using negative scores for sorting allows us to sort in ascending order and directly pick the top `k` students without reversing the order later.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Sorting, Heap (Priority Queue).

Companies

Asked at: Booking.com.

Reward Top K Students | LeetCode 2512

Medium

Array Hash Table String Sorting Heap (Priority Queue)

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<Integer> topStudents(String[] positive_feedback, String[] negative_feedback, String[] report, int[] student_id, int k) {
        Set<String> pos_words = new HashSet<>(Arrays.asList(positive_feedback));
        Set<String> neg_words = new HashSet<>(Arrays.asList(negative_feedback));
        
        List<Pair> student_info = new ArrayList<>();
        for (int i = 0; i < student_id.length; i++) {
            int sid = student_id[i];
            String rep = report[i];
            int score = 0;
            for (String word : rep.split(" ")) {
                if (pos_words.contains(word)) {
                    score += 3;
                } else if (neg_words.contains(word)) {
                    score -= 1;
                }
            }
            student_info.add(new Pair(-score, sid));
        }
        
        Collections.sort(student_info);
        
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < k && i < student_info.size(); i++) {
            result.add(student_info.get(i).sid);
        }
        
        return result;
    }

    static class Pair implements Comparable<Pair> {
        int score;
        int sid;

        public Pair(int score, int sid) {
            this.score = score;
            this.sid = sid;
        }

        @Override
        public int compareTo(Pair other) {
            return Integer.compare(this.score, other.score);
        }
    }
}

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Problem Info

Description

Problem 2512: Reward Top K Students - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting Heap (Priority Queue)

Companies

Booking.com

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2512

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Create hash sets (or sets) for positive and negative feedback words for efficient lookup.
2
Step 2: Iterate through each student's report and calculate their score. For each word in the report, check if it's in the positive or negative feedback set and update the score accordingly.
3
Step 3: Store each student's score and ID as a pair in a list.
4
Step 4: Sort the list of student information based on the score in descending order (using a negative score for simpler sorting) and then by student ID in ascending order (for tie-breaking).
5
Step 5: Extract the top `k` student IDs from the sorted list and return them.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting step, while the space complexity depends on the size of the data structures used to store the student information and the feedback words.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Heap (Priority Queue)

Use a min-heap of size `k` to maintain the top `k` students seen so far. For each student, calculate the score. If the score is greater than the minimum score in the heap, or if the heap has fewer than `k` elements, add the student to the heap and adjust. This approach is better when k is relatively small compared to n.

Time: O(n log k), where n is the number of students. Inserting into the heap takes O(log k) time, and this is done for each of the n students.Space: O(k), as the heap stores at most `k` student entries.

Trade-offs: The heap-based approach is better when `k` is significantly smaller than `n`, as O(n log k) can be more efficient than O(n log n). However, it's more complex to implement than the sorting-based approach.

✨

Key Insights

1Insight 1: Representing positive and negative words efficiently using hash sets (or sets in Python) allows for quick lookup when calculating the score for each student.
2Insight 2: Storing student information (score and ID) as a pair (or a similar data structure) is crucial for sorting them based on score and ID.
3Insight 3: Using negative scores for sorting allows us to sort in ascending order and directly pick the top `k` students without reversing the order later.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When k is greater than the number of students, the solution correctly returns all the student IDs. The `min` function or a similar check is implicitly handled by the loop condition in some implementations.
Edge case 2: Empty feedback reports. The code handles this gracefully as the score remains 0 for the student in this scenario. The loops iterate 0 times when the report is empty. Tokenization of an empty string will result in no tokens.
Edge case 3: When all students have the same score. The code correctly sorts them by their IDs, ensuring the student with the lower ID ranks higher.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the score by not distinguishing between positive and negative feedback or by not using sets for efficient word lookup.

✗Common mistake 2: Not handling the tie-breaking condition correctly, leading to incorrect ranking when students have the same score.

✗Common mistake 3: Not handling the case when k is larger than the number of students, resulting in an out-of-bounds error or incorrect output.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Ranking products based on customer reviews. Positive reviews increase the product's score, while negative reviews decrease it. The top-ranked products are then displayed to users.

🏦Real-world scenario 2: Identifying top-performing employees based on performance reviews. Positive feedback from managers and peers increases the employee's score, while negative feedback decreases it. The top-performing employees are then considered for promotions or rewards.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reward Top K Students"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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